How would I go about solving for $x$?

Question

How would I approach this problem? (Solving for $x$) $$x^{x}=e^{\Omega}$$ I tried using logarithms and rearranging, but it didn't seem to help: $$x=e^{\frac{\Omega}{x}}$$ $$\ln(x)=\frac{\Omega}{x}$$ $$x\cdot\ln(x)=\Omega$$ Where do I go from here? Is this a bad approach?

Answer 1

The Lambert W function comes in handy, just set the equation up so that $x=W(x)e^{W(x)}$ can be used. $$x^x=e^{\Omega}$$ $$\Rightarrow x=e^{\frac{\Omega}{x}}$$ $$\Rightarrow \Omega=\frac{\Omega}{x}\cdot e^{\frac{\Omega}{x}}$$ Now you can apply the Lambert W function: $$W(\Omega)=\frac{\Omega}{x}$$ $$\Rightarrow x=\frac{\Omega}{W(\Omega)}$$ This can still be simplified by rearranging $x=W(x)e^{W(x)}\Rightarrow e^{W(x)}=\frac{x}{W(x)}$. $$\boxed{x=e^{W(\Omega)}\approx1.47058}$$